Problem is:
3 prizes, one for English, one for French and one for Spanish can be awarded in a class of 20 students. Find the number of different ways in which the three prizes can be awarded if:
(i) no student may win > 1 prize.
(ii) no student may win all 3 prizes.

But I am struggling with (ii). If the question asked was student CAN win 1,2 or 3 prizes then poss ways would be 20*20*20=8000. Is that correct.

My first thought was that way to do this is 20*20*19 - but that is incorrect. Correct way to solve is apparently 20*20*20 - 20. But why?

Angus

May 8th 2011, 04:21 AM

Plato

Quote:

Originally Posted by angypangy

3 prizes, one for English, one for French and one for Spanish can be awarded in a class of 20 students. Find the number of different ways in which the three prizes can be awarded if:
(ii) no student may win all 3 prizes.
My first thought was that way to do this is 20*20*19 - but that is incorrect. Correct way to solve is apparently 20*20*20 - 20. But why?

There are functions from a set of three to a set of twenty. But there are 20 ways that all three go to the same image (in this case a student). So subtract..

May 8th 2011, 04:26 AM

angypangy

Sorry to be a pain but how do you calculate the 20. you say 20 ways that all three go to the same image? How are you getting the 20 to subtract?

May 8th 2011, 04:39 AM

angypangy

Its ok I have got it now.

May 8th 2011, 04:39 AM

Plato

Quote:

Originally Posted by angypangy

Sorry to be a pain but how do you calculate the 20. you say 20 ways that all three go to the same image? How are you getting the 20 to subtract?

How many ways can you give all three prizes to the same student?
Those ane the cases you want to exclude (no student gets all three).